传送门:【CodeForces】343D Water Tree
题目分析:首先dfs求出dfs序,以此作为线段树的坐标。操作1就是将区间[ in[ v ] , ou[ v ] ]内的数都变为1,操作2就是将位置in[ v ]上的数变为0,操作3即查询区间[ in[ v ] , ou[ v ] ]内是否有0存在。因为如果子树内存在一个0,则这个节点一定会变为0。操作1还要注意一点就是如果子树内存在0,那么v的父节点就应该更新为0,因为将[ in[ v ] , ou[ v ] ]更新为1后v的祖先们不会改变(因为一个点是否为0看的是他所代表的区间内是否有0),这样才能保证算法的正确性。
代码如下:
#include <cmath> #include <cstdio> #include <cstring> #include <algorithm> using namespace std ; #define REP( i , a , b ) for ( int i = ( a ) ; i < ( b ) ; ++ i ) #define FOR( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i ) #define REV( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i ) #define travel( e , H , u ) for ( Edge* e = H[u] ; e ; e = e -> next ) #define CLR( a , x ) memset ( a , x , sizeof a ) #define ls ( o << 1 ) #define rs ( o << 1 | 1 ) #define lson ls , l , m #define rson rs , m + 1 , r #define root 1 , 1 , n #define rt o , l , r #define mid ( ( l + r ) >> 1 ) const int MAXN = 500005 ; const int MAXE = 1000005 ; struct Edge { int v ; Edge* next ; } E[MAXE] , *H[MAXN] , *cur ; int num[MAXN << 2] , set[MAXN << 2] ; int in[MAXN] , ou[MAXN] , dfs_clock ; int fa[MAXN] ; int n , q ; void clear () { cur = E ; dfs_clock = 0 ; CLR ( H , 0 ) ; CLR ( in , 0 ) ; CLR ( ou , 0 ) ; CLR ( num , 0 ) ; CLR ( set , 0 ) ; } void addedge ( int u , int v ) { cur -> v = v ; cur -> next = H[u] ; H[u] = cur ++ ; } void dfs ( int u , int f ) { fa[u] = f ; in[u] = ++ dfs_clock ; travel ( e , H , u ) if ( e -> v != f ) dfs ( e -> v , u ) ; ou[u] = dfs_clock ; } void pushup ( int o ) { num[o] = min ( num[ls] , num[rs] ) ; } void pushdown ( int o ) { if ( set[o] ) { set[ls] = set[rs] = set[o] ; num[ls] = num[rs] = set[o] ; set[o] = 0 ; } } void update ( int L , int R , int v , int o , int l , int r ) { if ( L <= l && r <= R ) { set[o] = num[o] = v ; return ; } pushdown ( o ) ; int m = mid ; if ( L <= m ) update ( L , R , v , lson ) ; if ( m < R ) update ( L , R , v , rson ) ; pushup ( o ) ; } int query ( int L , int R , int o , int l , int r ) { if ( L <= l && r <= R ) return num[o] ; pushdown ( o ) ; int m = mid ; if ( R <= m ) return query ( L , R , lson ) ; if ( m < L ) return query ( L , R , rson ) ; return min ( query ( L , R , lson ) , query ( L , R , rson ) ) ; } void solve () { int u , v ; clear () ; REP ( i , 1 , n ) { scanf ( "%d%d" , &u , &v ) ; addedge ( u , v ) ; addedge ( v , u ) ; } dfs ( 1 , 0 ) ; scanf ( "%d" , &q ) ; while ( q -- ) { scanf ( "%d%d" , &u , &v ) ; //printf ( "%d %d\n" , in[v] , ou[v] ) ; if ( u == 1 ) { if ( !query ( in[v] , ou[v] , root ) && fa[v] ) update ( in[fa[v]] , in[fa[v]] , 0 , root ) ; update ( in[v] , ou[v] , 1 , root ) ; } else if ( u == 2 ) update ( in[v] , in[v] , 0 , root ) ; else printf ( "%d\n" , query ( in[v] , ou[v] , root ) ) ; } } int main () { while ( ~scanf ( "%d" , &n ) ) solve () ; return 0 ; }